Little Math Help Please

[Range Me]

..well my teacher is crazy and i have no fucking clue what im doing wrong...

ax+x-5a-5
__________=divided by*

ax-x-5a+5

You have to reduce to lowest terms...

I got as far as this..

x(a+1)-5(a-1)
____________

x(a-1)-5(a+1)


Like i thought i could just cancel out like terms but it doesn't work..

This is college algebra and yea...ty if someone can explain how to do this...test is thrusday =[

 

[The Fat Controller]

Try multiplying the entire fraction by the denominator divided by the denominator, factorize the top and bottom, and see what you can cancel out.

Don't know if that is the right method though, it's 3am here and I'm rusty

 

[trainin4you]

the fat controllers idea sounds right tell me if it isnt and ill help ya

 

[Koot]

Ah I got it.

First factor out the a from top and bottom. Then you can factor out x-5 from the top. In order to do the same from the bottom you have to multiply by -1 to switch it around. After that is all done you can cancel out some terms.

In the end I got (a+1)/(a-1)

 

[MatthewGor123]

..well my teacher is crazy and i have no fucking clue what im doing wrong...

ax+x-5a-5
__________=divided by*

ax-x-5a+5

You have to reduce to lowest terms...

I got as far as this..

x(a+1)-5(a-1)
____________

x(a-1)-5(a+1)


Like i thought i could just cancel out like terms but it doesn't work..

This is college algebra and yea...ty if someone can explain how to do this...test is thrusday =[

Fat, you wouldn't need to do that as you don't have a quadratic expression in the numerator or denominator.

Koot is correct, but I don't think his explanation is very accurate.

You factored incorrectly in your very first step.

ax + x - 5a - 5 = (a+1)x - 5(a+1), not (a-1). That's because when you distribute a negative (-5) over a positive (+1), you end up with -5, which is what you want.

The denominator was also incorrect;
ax - x - 5a + 5 = (a-1)x - 5(a-1), not (a+1), because when you distribute a negative (-5) over a negative (-1), you get a positive (+5).

So, you should have:
(a+1)x - 5(a+1)
(a-1)x - 5(a-1).

From here, we can do some further factoring -
(a+1)(x-5)
(a-1)(x-5)

If this step is unclear, you can rename (for the numerator) the expression (a+1) as z, and it should be more clear. If you have z*x - 5*z, which is factored as z(x-5). Then, you just plug back (a+1) for z, and this step makes sense. The same is done for the numerator.

Of course, you see (x-5) in the numerator and denominator, and those cancel out, leaving you simply:
(a+1)
(a-1)

In the future, if you have any homework questions at all, please don't hesitate to ask on my thread , which can be found here: http://sythe.org/showthread.php?t=490925

 

[d 7 x]

Oh boy, I'm only in geometry. I don't think I want to go to College Algebra now

 

[MatthewGor123]

Oh boy, I'm only in geometry. I don't think I want to go to College Algebra now

Why? This isn't that hard :O I had much more trouble in my honors geometry than honors algebra...

 

[Range Me]

Fat, you wouldn't need to do that as you don't have a quadratic expression in the numerator or denominator.

Koot is correct, but I don't think his explanation is very accurate.

You factored incorrectly in your very first step.

ax + x - 5a - 5 = (a+1)x - 5(a+1), not (a-1). That's because when you distribute a negative (-5) over a positive (+1), you end up with -5, which is what you want.

The denominator was also incorrect;
ax - x - 5a + 5 = (a-1)x - 5(a-1), not (a+1), because when you distribute a negative (-5) over a negative (-1), you get a positive (+5).

So, you should have:
(a+1)x - 5(a+1)
(a-1)x - 5(a-1).

From here, we can do some further factoring -
(a+1)(x-5)
(a-1)(x-5)

If this step is unclear, you can rename (for the numerator) the expression (a+1) as z, and it should be more clear. If you have z*x - 5*z, which is factored as z(x-5). Then, you just plug back (a+1) for z, and this step makes sense. The same is done for the numerator.

Of course, you see (x-5) in the numerator and denominator, and those cancel out, leaving you simply:
(a+1)
(a-1)

In the future, if you have any homework questions at all, please don't hesitate to ask on my thread , which can be found here: http://sythe.org/showthread.php?t=490925

wow tyvm ..also ty to koot..but im still confused on like last step..

So, you should have:
(a+1)x - 5(a+1)
(a-1)x - 5(a-1).

From here, we can do some further factoring -
(a+1)(x-5)
(a-1)(x-5)

how did you get to (a+1)(x-5)???

 

[MatthewGor123]

wow tyvm ..also ty to koot..but im still confused on like last step..

So, you should have:
(a+1)x - 5(a+1)
(a-1)x - 5(a-1).

From here, we can do some further factoring -
(a+1)(x-5)
(a-1)(x-5)

how did you get to (a+1)(x-5)???

If this step is unclear, you can rename (for the numerator) the expression (a+1) as z, and it should be more clear. If you have z*x - 5*z, which is factored as z(x-5). Then, you just plug back (a+1) for z, and this step makes sense. The same is done for the denominator.

See what happens when you expand (a+1)(x-5) - by FOIL, you get ax - 5a + x - 5, which is what we were looking for. This is because (since multiplication is commutative; a*b = b*a) therefore (a+1)x - 5(a+1) can be rewritten as x(a+1) - 5(a+1), so a+1 is a factor in both of these expressions. What would xy - 5y be equal to? y(x-5). (a+1) is basically y in this. I'll explain it again if you're still unclear. If you add me on MSN, I can explain it in a bit more detail if it still doesn't make sense.

nerds..

Meh, when I'm driving my dream car and you're pumping my gas, we'll see who's laughing. How can you call me a nerd, if you've never even met me? Spam some more, dickhead.

 

[Range Me]

See what happens when you expand (a+1)(x-5) - by FOIL, you get ax - 5a + x - 5, which is what we were looking for. This is because (since multiplication is commutative; a*b = b*a) therefore (a+1)x - 5(a+1) can be rewritten as x(a+1) - 5(a+1), so a+1 is a factor in both of these expressions. What would xy - 5y be equal to? y(x-5). (a+1) is basically y in this. I'll explain it again if you're still unclear. If you add me on MSN, I can explain it in a bit more detail if it still doesn't make sense.



Meh, when I'm driving my dream car and you're pumping my gas, we'll see who's laughing. How can you call me a nerd, if you've never even met me? Spam some more, dickhead.

i understand when you FOIL the (a-1)(x-5) you get the original step before..but i don't understand how you canceled some out..

So, you should have:
(a+1)x - 5(a+1)
(a-1)x - 5(a-1).

From here, we can do some further factoring -
(a+1)(x-5)
(a-1)(x-5)

like x-5's can cancel out..so that should leave you with...

(a+1)(a+1) over (a-1)(x-1)

 

[Shin]

Well, I cannot understand a thing you did with this problem.

I wrote it down and did it in a few seconds.

So you start off with the initial problem:

ax + x - 5a - 5
----------------
ax - x - 5a - 5

From here you will get the following:

x(a+1) - 5(a+1)
----------------
x(a-1) - 5(a-1)

As a rule, you can do the following.

When you have AX + BX, it equals X(A+B)

So then... you can group the terms to the following

(x-5)(a+1)
-----------
(x-5)(a-1)

From here, simply cancel out the "(x-5)" from the Numerator / Denominator

Your final answer:

(a+1)
------
(a-1)

NOTE: I did this problem rather quickly, and may have made a mistake.
I'm in Pre-Cal currently, and will be in AP Calculus AB next Semester.

Any other Math questions, please PM me. Thanks.

 

[MatthewGor123]

i understand when you FOIL the (a-1)(x-5) you get the original step before..but i don't understand how you canceled some out..

So, you should have:
(a+1)x - 5(a+1)
(a-1)x - 5(a-1).

From here, we can do some further factoring -
(a+1)(x-5)
(a-1)(x-5)

like x-5's can cancel out..so that should leave you with...

(a+1)(a+1) over (a-1)(x-1)

Okay, so you're asking how we got from
(a+1)(x-5)
(a-1)(x-5)

To just:
(a+1)
(a-1)
I'm assuming?

So let's say you have (x-5) is equal to some arbitrary variable z.
(a+1)(z)
(a-1)(z)

If you have, let's say, a = 9 --> totally arbitrary, this is just so you can visualize it better, you would have:
10 * z
8 * z

Regardless of whatever z is, 10*z and 8*z will always have z as a factor, therefore allowing us to cancel it out when we're simplifying. For example, if z = 5, does you can simplify 50/40 to 10/8 [and even further, but that's beside the point]. Does it make sense up to here?

If it does, it should make sense that since (x-5) is in parentheses, it can be viewed as a separate variable - z, where x-5 = z. Basically, you would have:

(a+1) (x-5)
----- x -----
(a-1) (x-5)

And, as you know,
(x-5)
----- = 1
(x-5)

So therefore, you're just multiplying:
(a+1)
(a-1) by one, which of course does not change the answer.

Hopefully this makes a bit more sense. I think it's about as much detail as I can explain it in, without using live-chat. Please MSN me if you're still unclear

 

[MatthewGor123]

Well, I cannot understand a thing you did with this problem. I wrote it down and did it in a few seconds.

Uhh, I answered it yesterday...I was just trying to explain it - of course, this is the correct method, and the method I used...He's a bit unclear with the algebra rules, so I was just trying to explain it...Jeez, no need to be rude about it...

[EDIT: Sorry about double post]

 

[Shin]

Uhh, I answered it yesterday...I was just trying to explain it - of course, this is the correct method, and the method I used...He's a bit unclear with the algebra rules, so I was just trying to explain it...Jeez, no need to be rude about it...

[EDIT: Sorry about double post]

For one, it was not an accident for the double post.
It was intentional because you didn't want to edit your other thread.

And I wasn't being rude about it. It's just that you have numbers everywhere.

I simply posted in response a quick way of explaining how to solve the problem. It is explained in a quick and efficient fashion.

*EDIT*
And don't reply in some sort of attack, because it's merely spamming more.

This thread was made to ask about help for a Math Problem, and that is all. The problem has been solved.

 

[MatthewGor123]

For one, it was not an accident for the double post.
It was intentional because you didn't want to edit your other thread.

And I wasn't being rude about it. It's just that you have numbers everywhere.

I simply posted in response a quick way of explaining how to solve the problem. It is explained in a quick and efficient fashion.

*EDIT*
And don't reply in some sort of attack, because it's merely spamming more.

This thread was made to ask about help for a Math Problem, and that is all. The problem has been solved.

No, it was an accident. I thought you had posted after me, I didn't realize until after that your post was before mine. My thread has numbers everywhere? He said "his teacher is crazy, he doesn't understand it" so I assumed that he was looking for a LOGICAL, REASONABLE explanation, rather than one based on memorizing rules. I explained to him an easy way of remembering that Ax + Bx = (A+B)x. My steps were the exact same as yours, but I included an explanation in between. Because someone posted before both of us with the correct answer, and an almost identical level of explanation.

And obviously my explanation was in more detail than yours, but he still didn't fully understand it. But honestly, all I care about is if ultimately he understands the process ^_^

And how are we spamming? Lol. These messages are very relevant to the topic; we are trying to make his understanding of the problem better.

 

[Range Me]

noo i underatand how to cross them out..its the step before that...

(a+1)x - 5(a+1)
(a-1)x - 5(a-1).


how do you get from that above to

the next step....

(a+1)(x-5)
(a-1)(x-5)

...like on the first part where does the a+1 and a-1 go?

 

[MatthewGor123]

noo i underatand how to cross them out..its the step before that...

(a+1)x - 5(a+1)
(a-1)x - 5(a-1).


how do you get from that above to

the next step....

(a+1)(x-5)
(a-1)(x-5)

...like on the first part where does the a+1 and a-1 go?

Add me on MSN mate, I think it's been explained as much as it can be.

(a+1)x - 5(a+1) is the numerator. The (a+1) does not go anywhere, that whole expression is just written in simplified form. (a+1)(x-5) is just a simplified version of that (a+1)x - 5(a+1). This is because, as Unbanned said, for any variable "V", aV + bV = (a+b)V. In this case, V = (a+1), a = x, and b=(-5), therefore (x-5)(a+1) is just the simplified form of that. It doesn't go anywhere, it's just simplified.

Please add [email protected] on MSN.

 

[The Fat Controller]

I did some doodling on MS paint since I was bored Hope this helps or at least irritates somebody:

 

[MatthewGor123]

Exactly Already explained on MSN though ^_^ The guy's accusing me of spamming...all I'm trying to do is fucking help.

 

[Range Me]

Exactly Already explained on MSN though ^_^ The guy's accusing me of spamming...all I'm trying to do is fucking help.

Don't listen to those fag's you helped me so much ty.